Hey Divya

Solution (S):

S = 0.7+0.77+0.777+....0.7777

=7(0.1+0.11+0.11+0.111+.......+0.1111....)

= 79[0.9+0.99+0.99+0.999+...........)

=79[(1−0.1)+(1−0.01)+....+(1−0.000......1)]

=7/9 * [20 − (1 / 10+1 / 10^2+......1 / 20^20)]

Therefore, a= r = 1/10

=7/ 9* [20−1/9 * (1−10^−20)

=7 / 81 * [179 + 10^−20]  -> Answer

Hello!!!

Hope you are doing great!!!

Given series = 1+3/2+5/4+7/8.........

=1+3/2+5/4+7/8+9/16+11/32+..........

multiply and divide by 2

=2(1+1/2+1/4+1/8+1/16+1/32+............)

=2+2(1)+2(1/2)+2(1/4)+2(1/8)+2(1/16)+..................

2+2(1+1/2+1/4+1/8+1/16+..............)

here (1+1/2+1/4+1/8+1/16+..............) is in GP

The value of (1+1/2+1/4+1/8+1/16+..............)=2

substitute in the above equation

=2+2(2)

=2+4

=6

Hope it helps you!!

Thank you!!!

You have almost 1 year to appear for JEE main January exam you can follow the following tips to do well in the exam.

• If you feel your physics is weak then start by developing concepts it is a proven fact that most of the aspirants start by following HC

Hello Subhodeep

There are several books, you may consider any of the below mentioned:

Organic Chemistry by O P Tandon, NCERT, NOOTAN Chemistry, etc.

For more list of good and recommended books for JEE please refer the below mentioned link:

https://engineering.careers360.com/articles/best-books-for-jee-main

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I understand what you are trying to ask and that can only be achieved with consistent solving questions. Nothing can help you to improve your speed  but your practice to solve mcqs now . Set a clock with timer of one hour or 30 mins . Take

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